Question

# A coffee cup calorimeter initially contains 125g of water at a temperature of${{24.2}^{\circ }}C$. After adding 10.5 g of KBr, the temperature becomes ${{21.1}^{\circ }}C$. The heat of solution is:A. 40 J/gB. 117 J/gC. 167.7 J/gD. 420.05 J/g

Hint: Heat of solution is also called as the enthalpy of solution, which is the enthalpy change in relation with the dissolution of a solute in a solvent at a constant pressure, that results in infinite dilution. It has a unit of KJ/mol.

- We know that the heat energy released is equal to
\begin{align} & 125\times (24.2-21.1) \\ & =387.5\text{ }cal \\ \end{align}
- Now the heat of solution =$\dfrac{heat\text{ }energy\text{ }released}{weight\text{ }of\text{ }solute}$
$\dfrac{420.05}{10.5}ca\operatorname{l}\text{ }{{g}^{-1}}$
=40cal/g
- So, one calories are equal to 4.126J,
Hence, heat of solution will be
\begin{align} & 40\times 4.126\text{ }J/g \\ & =167\text{ }J/g \\ \end{align}
- There, we can conclude that the correct option is (c), that is the heat of solution is 167.7 J/g.